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Appendix II.


exist alone, for they cause a turning eifect. They must, therefore,
be balanced by a second pair of stresses F2F2, where Fxx AB =
F2 x A D, or (/! A D) A B = (f2 A B) A D, and^ = /2.

Let us next imagine a second block A B c D, Fig. 822, acted
on by a shear stress /s and a direct stress ft. These may be
balanced by the dotted stresses, but we shall consider them
resisted by f& and f0 on plane c B. Now if the value of d be
properly chosen, f0 may be entirely eliminated, and the forces

on B c be solely, direct, as /e.     Assuming  this  condition,   and
resolving   ft c B   on c A :

(/eCB)sin 0 = (/tAB)-(/s CA)
Dividing by c B :       /*. sin 6 = ft sin 0 - f$ cos 6

and (/e -ft) sin 6 =  -/s cos 6.....................    (i)

Resolving (/i c B) on A B :

(ft c B) cos B =  - /s A B
Dividing by c B :   y^ cos 0 =   j^sinO   ............    (2)

Multiplying (i) and (2) together:   /e2 -/e/t = /s2    '

and solving the quadratic :      /e  J (/t 4- V/t2 + 4/s2)
Inserting the values ofj^ andj^ on p. 461: